# Using Conversion Factors

In Europe, gasoline efficiency is measured in $\mathrm{km/L}$. If your car's gas mileage is $35.0~\mathrm{mi/gal}$, how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions: $1~\mathrm{km} = 0.6214~\mathrm{mi}$ and $1~\mathrm{gal} = 3.78~\mathrm{L}$. Express your answer numerically in liters.

I am bit confused as to how to compute my last step.

My conversion pathway:

$$\frac{35.0\ \textrm{mi}}{1\ \textrm{gal}}×\frac{1\ \textrm{km}}{0.6214\ \textrm{ mi}}×\frac{1\ \textrm{gal}}{3.78\ \textrm{L}}= 15\ \textrm{km/L}$$ So: $$15\ \textrm{km/L}×142/\textrm{km}=2115\ \textrm{km}$$

Am I right?

• Note that Europe generally uses litres/100 km and not a mileage per litre. – Jan Oct 30 '15 at 17:18

You did the conversion to km/L right, but I wouldn't round to 15 right away. In your last step you multiplied, should have divided, 142 km / 15 km/L = X L. Just keep track of units and that will help you determine how to do the math.

• So I should round the number and determine the significant figures when I have my final calculation? – Cetshwayo Sep 12 '14 at 18:34
• Why am I dividing 142 by 15? Please explain? – Cetshwayo Sep 12 '14 at 18:40
• In general, don't round until the end, that will help preserve accuracy. And you divide 142 km by 15km/L because you want L as your answer. Dividing km by km/L is the same as multiplying km by L/km. This puts a km on each side of the fraction and they cancel out, leaving L. – user137 Sep 12 '14 at 18:55